At 3:30 p.m., Berto’s train was 34 miles past the egg farm, traveling at an average speed of 85 miles per hour. At the same time on a nearby track, Eduardo’s train was traveling at an average speed of 110 miles per hour and had 12 miles to go before it reached the egg farm. To the nearest hundredth of an hour, after how much time will the trains meet up?
+20 There is a 46 miles difference between the trains (34+12). Eduardo's train travels 25 miles faster than Berto's train (110-85). After one hour there should only be 21 miles difference between the trains (46-25). It takes 84% of an hour for Eduardo's train to catch up (21*100/25). 84*60/100 = 50.4 minutes, which is 50 minutes and 24 seconds. So added up it takes 1 hour 50 minutes and 24 seconds until the trains meet up, which is at 5:20:24 p.m.
Hopefully I didn't make any mistakes while calculating using miles, since I am from the Netherlands and we use kilometers here.
Well anyway, I hope this helps you out ^w^
+20 There is a 46 miles difference between the trains (34+12). Eduardo's train travels 25 miles faster than Berto's train (110-85). After one hour there should only be 21 miles difference between the trains (46-25). It takes 84% of an hour for Eduardo's train to catch up (21*100/25). 84*60/100 = 50.4 minutes, which is 50 minutes and 24 seconds. So added up it takes 1 hour 50 minutes and 24 seconds until the trains meet up, which is at 5:20:24 p.m.
Hopefully I didn't make any mistakes while calculating using miles, since I am from the Netherlands and we use kilometers here.
Well anyway, I hope this helps you out ^w^
Speed =Distance/Time
Let the distance travelled before the 2 trains meet=D, then
D/85 =[D+46]/110, solve for D
D =156.4 miles - when the 2 trains will meet.
156.4/85 =1.84 hours- after 3:30 when they will meet, or 110.4 minutes after 3:30pm. So that:
3:30 + 1.84 =5:20:24pm - when the 2 trains will meet.
